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Let $a_n\geq 0$. $\sum_{n=1}^\infty \frac{a_1+a_2+\cdots+a_n}{n}$ converges/divereges? Take $a_n=1/n$ for example, then the proposed series looks like $\sum\frac{\ln n}{n}$, which is divergent. What about the general case?

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  • $\begingroup$ it cud also converge obviously; e.g. $a_n = 0$ for each $n$ $\endgroup$ – mathworker21 Jun 9 '19 at 12:39
  • $\begingroup$ Or even worse: $\frac{1+2+3+\cdots+n}{n}$ clearly diverges. You need an extra condition such as $\lim_{n \to \infty} a_n = 0$ or that $a_n$ is decreasing. $\endgroup$ – Toby Mak Jun 9 '19 at 12:40
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    $\begingroup$ If $a_1 > 0$ then the sum is at least $\sum_{n = 1}^{\infty}\tfrac{a_1}{n}$ which diverges. Similarly if any of the $a_n$'s are positive, the sum diverges. $\endgroup$ – JimmyK4542 Jun 9 '19 at 12:41
  • $\begingroup$ looks like cauchy's first theorem on limit. see here math.stackexchange.com/questions/1930373/… $\endgroup$ – aud098 Jan 24 '20 at 16:33
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Convergence holds only when $a_n=0$ for all $n$. $\sum\limits_{n=1}^{\infty} \frac 1 n \sum\limits_{i=1}^{n} a_i=\sum\limits_{i=1}^{\infty}a_i \sum\limits_{n=i}^{\infty} \frac 1 n=\infty$ unless $a_i=0$ for all $i$.

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  • $\begingroup$ [ I have used the fact that $\sum\limits_{n=i}^{\infty} \frac 1 n = \infty$ for every $i$]. $\endgroup$ – Kavi Rama Murthy Jun 9 '19 at 12:48
  • $\begingroup$ Note that the interchange in order of summation is justified by Tonelli's theorem. $\endgroup$ – Math1000 Jun 9 '19 at 12:55

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